subring of rational numbers and its ideal

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Change of notation: let me call $\mathbb Z_{(p)}$ what you call $\mathbb Q_p$. It is the localization of $\mathbb Z$ at the multiplicative subset $S=\mathbb Z\setminus p\mathbb Z$. All that lies in $S$ becomes (universally) invertible in $\mathbb Z_{(p)}$ (and we formally write these elements as denominators...). So we have a chain of ring homomorphisms $$\mathbb Z\hookrightarrow \mathbb Z_{(p)}\hookrightarrow\mathbb Q.$$

That said, we have $$ P=\Bigl\{\frac{a}{b}\,:\,a\notin S\Bigr\}=\Bigl\{\frac{a}{b}\,:\,a\in p\mathbb Z\Bigr\}=p\mathbb Z_{(p)}. $$ In other words, $P$ is the ideal generated by $p$, or by $\frac{p}{1}$ if we want to be precise. As $P$ was defined to be $\mathbb Z_{(p)}\setminus \{\textrm{units}\}$, we have that $$\{\textrm{units}\}=\mathbb Z_{(p)}^\times=\mathbb Z_{(p)}\setminus p\mathbb Z_{(p)}.$$ This implies that $p\mathbb Z_{(p)}$ is the unique maximal ideal. Finally, let us follow the hint and define a map $$\phi:\mathbb Z_{(p)}\to \mathbb Z/p\mathbb Z$$ by $\phi(a/b)=\overline a.\overline b^{-1}$. Then $$\ker\phi=\Bigl\{\frac{a}{b}:\overline a.\overline b^{-1}=0_{Z/p\mathbb Z}\Bigr\}=\Bigl\{\frac{a}{b}:\overline a=0_{Z/p\mathbb Z}\Bigr\}=\Bigl\{\frac{a}{b}:p\,\textrm{divides}\,a\Bigr\}=P,$$ where we just used that the inverse of $\overline b$ cannot be zero in $Z/p\mathbb Z$.

Aside: in your question you say that $P$ is a subring. It cannot be a subring, as a proper ideal is never a subring.

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Updated on August 01, 2022

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  • Nika
    Nika over 1 year

    Let $p$ be a prime number. For any $p$ the subring $\mathbb{Q}_p$ of of the field of rational numbers is defined:

    $\mathbb{Q}_p=\{\frac{a}{b}|a,b\mbox{ are integers, $p$ does not divide $b$}\}$

    Let $P$ be a subring of $\mathbb{Q}_p$ that is the set of all elements that are NOT invertible in $\mathbb{Q}_p$.

    i) Show that $P$ is an ideal in $\mathbb{Q}_p$.

    ii) Show that $P$ is a unique maximal ideal in $\mathbb{Q}_p$.

    iii) Prove that the quotient ring $\mathbb{Q}_p/P$ is isomorphic to the field $\mathbb{Z}_p$.

    There was given a hint that one should examine a map $\phi$ s.t. $\phi\left(\frac{a}{b}\right)=\bar{a}\cdot\left(\bar{b}\right)^{-1}$; the last is mod $p$.